How do you find int ( x + 1)/(x^3 + x^2 -2x) dx using partial fractions?

Mar 13, 2018

The answer is $= - \frac{1}{2} \ln \left(| x |\right) - \frac{1}{6} \ln \left(| x + 2 |\right) + \frac{2}{3} \ln \left(| x - 1 |\right) + C$

Explanation:

Perform the decomposition into partial fractions

$\frac{x + 1}{{x}^{3} + {x}^{2} - 2 x} = \frac{x + 1}{x \left({x}^{2} + x - 2\right)}$

$= \frac{x + 1}{x \left(x + 2\right) \left(x - 1\right)}$

$= \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 1}$

$= \frac{A \left(x + 2\right) \left(x - 1\right) + B \left(x\right) \left(x - 1\right) + C \left(x\right) \left(x + 2\right)}{x \left(x + 2\right) \left(x - 1\right)}$

The denominators are the same, compare the numerators

$x + 1 = A \left(x + 2\right) \left(x - 1\right) + B \left(x\right) \left(x - 1\right) + C \left(x\right) \left(x + 2\right)$

Let $x = 0$, $\implies$, $1 = - 2 A$, $\implies$, $A = - \frac{1}{2}$

Let $x = - 2$, $\implies$, $- 1 = 6 B$, $\implies$, $B = - \frac{1}{6}$

Let $x = 1$, $\implies$, $2 = 3 C$, $\implies$, $C = \frac{2}{3}$

Therefore,

$\frac{x + 1}{{x}^{3} + {x}^{2} - 2 x} = \frac{- \frac{1}{2}}{x} + \frac{- \frac{1}{6}}{x + 2} + \frac{\frac{2}{3}}{x - 1}$

$\int \frac{\left(x + 1\right) \mathrm{dx}}{{x}^{3} + {x}^{2} - 2 x} = \int \frac{- \frac{1}{2} \mathrm{dx}}{x} + \int \frac{- \frac{1}{6} \mathrm{dx}}{x + 2} + \int \frac{\frac{2}{3} \mathrm{dx}}{x - 1}$

$= - \frac{1}{2} \ln \left(| x |\right) - \frac{1}{6} \ln \left(| x + 2 |\right) + \frac{2}{3} \ln \left(| x - 1 |\right) + C$